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        <p><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/parallel-courses-ii/">1494. 并行课程 II</a>给你一个整数 <code>n</code> 表示某所大学里课程的数目，编号为 <code>1</code> 到 <code>n</code> ，数组 <code>dependencies</code> 中， <code>dependencies[i] = [xi, yi]</code> 表示一个先修课的关系，也就是课程 <code>xi</code> 必须在课程 <code>yi</code> 之前上。同时你还有一个整数 <code>k</code> 。</p>
<p>在一个学期中，你 <strong>最多</strong> 可以同时上 <code>k</code> 门课，前提是这些课的先修课在之前的学期里已经上过了。</p>
<p>请你返回上完所有课最少需要多少个学期。题目保证一定存在一种上完所有课的方式。</p>
<p>嗯哼哼 会不会写呢</p>
<a id="more"></a>

<h1 id="位运算的一些技巧"><a href="#位运算的一些技巧" class="headerlink" title="位运算的一些技巧"></a>位运算的一些技巧</h1><h2 id="位运算的基本内容"><a href="#位运算的基本内容" class="headerlink" title="位运算的基本内容"></a>位运算的基本内容</h2><p>简单来说<br>位操作包括：</p>
<ul>
<li>~ 取反（NOT）</li>
<li>| 按位或（OR）</li>
<li>^ 按位异或（XOR）</li>
<li>&amp; 按位与（AND）</li>
<li>移位 (移位是一个二元运算符，用来将一个二进制数中的每一位全部都向一个方向移动指定位，溢出的部分将被舍弃，而空缺的部分填入一定的值。)<ul>
<li>算术移位</li>
<li>逻辑移位<br>这里简单解释下</li>
</ul>
</li>
</ul>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br></pre></td><td class="code"><pre><span class="line">a = <span class="number">21</span>           (<span class="number">1</span> <span class="number">0</span> <span class="number">1</span> <span class="number">0</span> <span class="number">1</span>)</span><br><span class="line">b = <span class="number">11</span>           (<span class="number">0</span> <span class="number">1</span> <span class="number">0</span> <span class="number">1</span> <span class="number">1</span>)</span><br><span class="line">c = ~a = <span class="number">-22</span>     (<span class="number">0</span> <span class="number">1</span> <span class="number">0</span> <span class="number">1</span> <span class="number">0</span>)</span><br><span class="line">d = a | b = <span class="number">31</span>   (<span class="number">1</span> <span class="number">1</span> <span class="number">1</span> <span class="number">1</span> <span class="number">1</span>)</span><br><span class="line">e = a ^ b = <span class="number">30</span>   (<span class="number">1</span> <span class="number">1</span> <span class="number">1</span> <span class="number">1</span> <span class="number">0</span>)</span><br><span class="line">f = a &amp; b = <span class="number">1</span>    (<span class="number">0</span> <span class="number">0</span> <span class="number">0</span> <span class="number">0</span> <span class="number">1</span>)</span><br><span class="line"><span class="number">1</span> &lt;&lt; <span class="number">4</span> = <span class="number">16</span>      (<span class="number">1</span> <span class="number">0</span> <span class="number">0</span> <span class="number">0</span> <span class="number">0</span>)</span><br><span class="line"><span class="number">16</span> &lt;&lt; <span class="number">2</span> = <span class="number">4</span>      (<span class="number">0</span> <span class="number">0</span> <span class="number">1</span> <span class="number">0</span> <span class="number">0</span>)</span><br></pre></td></tr></table></figure>

<h2 id="位运算的经典操作"><a href="#位运算的经典操作" class="headerlink" title="位运算的经典操作"></a>位运算的经典操作</h2><h3 id="简单的集合部分"><a href="#简单的集合部分" class="headerlink" title="简单的集合部分"></a>简单的集合部分</h3><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br></pre></td><td class="code"><pre><span class="line">A |=<span class="number">1</span> &lt;&lt; C     <span class="comment"># 在集合 A 中加入 C</span></span><br><span class="line">A &amp;= ~(<span class="number">1</span> &lt;&lt; C) <span class="comment"># 在集合 A 中删除 C </span></span><br><span class="line">A = <span class="number">1</span> &lt;&lt; C;    <span class="comment"># 相当于 A = 2 ** C</span></span><br><span class="line">a &amp; (-a)       <span class="comment"># lowbit 操作：获得某数的最后一位为 1 的数字</span></span><br><span class="line">A = <span class="number">0</span>          <span class="comment"># 将 A 置空</span></span><br><span class="line">All = (<span class="number">1</span> &lt;&lt; <span class="number">15</span>) - <span class="number">1</span>;  <span class="comment"># 将 All 置满</span></span><br><span class="line">(A &amp; B) == B   <span class="comment"># B 是否为 A 的子集</span></span><br></pre></td></tr></table></figure>

<h3 id="子集"><a href="#子集" class="headerlink" title="子集"></a>子集</h3><p>求排列组合是LeetCode周赛中常见的一个组成部分，如何快速简单的求排列组合呢：<br>在LeetCode<a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/number-of-valid-words-for-each-puzzle/">1178. 猜字谜</a>中有一个非常巧妙的方法：使用位运算。<br>这里的原始集合是一个原始单词的二进制位集合- ::单词中字母的数量不影响::，如下便是单词<code>abd</code>的二进制位集合。</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1</span><br><span class="line">z y x w v u t s r q p o n m l k j i h g f e d c b a</span><br></pre></td></tr></table></figure>

<p>如何求这个集合的子集呢，即所有字母都在原始单词中出现过的单词的二进制位集合。例如<code>be</code>、<code>a</code>等的二进制位集合分别是<code>0110</code>、<code>0001</code>。</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br></pre></td><td class="code"><pre><span class="line">subMask &#x3D; originalMask &#x3D; 1011 &#x2F;&#x2F; [d, b, a]</span><br><span class="line">subMask &#x3D; (subMask - 1) &amp; originalMask &#x3D; (1011 - 1) &amp; 1011 &#x3D; 1010 </span><br><span class="line">subMask &#x3D; (subMask - 1) &amp; originalMask &#x3D; (1010 - 1) &amp; 1011 &#x3D; 1001 </span><br><span class="line">subMask &#x3D; (subMask - 1) &amp; originalMask &#x3D; (1001 - 1) &amp; 1011 &#x3D; 1000 </span><br><span class="line">subMask &#x3D; (subMask - 1) &amp; originalMask &#x3D; (1000 - 1) &amp; 1011 &#x3D; 0011 </span><br><span class="line">subMask &#x3D; (subMask - 1) &amp; originalMask &#x3D; (0011 - 1) &amp; 1011 &#x3D; 0010</span><br><span class="line">subMask &#x3D; (subMask - 1) &amp; originalMask &#x3D; (0010 - 1) &amp; 1011 &#x3D; 0001</span><br><span class="line">subMask &#x3D; (subMask - 1) &amp; originalMask &#x3D; (0001 - 1) &amp; 1011 &#x3D; 0000</span><br></pre></td></tr></table></figure>

<p>简单用代码实现</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line">subMask = originalMask</span><br><span class="line">count = &#123;subMask&#125;</span><br><span class="line"><span class="keyword">while</span> subMask:</span><br><span class="line">    subMask = (subMask - <span class="number">1</span>) &amp; originalMask</span><br><span class="line">    count.add(submask)</span><br></pre></td></tr></table></figure>

<p>由此便优雅的完成子集的统计。</p>
<h3 id="1494-并行课程-II"><a href="#1494-并行课程-II" class="headerlink" title="1494. 并行课程 II"></a><a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/parallel-courses-ii/">1494. 并行课程 II</a></h3><p>给你一个整数 <code>n</code> 表示某所大学里课程的数目，编号为 <code>1</code> 到 <code>n</code> ，数组 <code>dependencies</code> 中， <code>dependencies[i] = [xi, yi]</code> 表示一个先修课的关系，也就是课程 <code>xi</code> 必须在课程 <code>yi</code> 之前上。同时你还有一个整数 <code>k</code> 。</p>
<p>在一个学期中，你 <strong>最多</strong> 可以同时上 <code>k</code> 门课，前提是这些课的先修课在之前的学期里已经上过了。</p>
<p>请你返回上完所有课最少需要多少个学期。题目保证一定存在一种上完所有课的方式。</p>
<p><strong>示例 1：</strong></p>
<p><strong><img src="https://tva1.sinaimg.cn/large/007S8ZIlly1gge2rhffrjj307h043jr9.jpg" alt="img"></strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入：n &#x3D; 4, dependencies &#x3D; [[2,1],[3,1],[1,4]], k &#x3D; 2</span><br><span class="line">输出：3 </span><br><span class="line">解释：上图展示了题目输入的图。在第一个学期中，我们可以上课程 2 和课程 3 。然后第二个学期上课程 1 ，第三个学期上课程 4 。</span><br></pre></td></tr></table></figure>

<p><strong>示例 2：</strong></p>
<p><strong><img src="https://tva1.sinaimg.cn/large/007S8ZIlly1gge2riauk8j307j05va9z.jpg" alt="img"></strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入：n &#x3D; 5, dependencies &#x3D; [[2,1],[3,1],[4,1],[1,5]], k &#x3D; 2</span><br><span class="line">输出：4 </span><br><span class="line">解释：上图展示了题目输入的图。一个最优方案是：第一学期上课程 2 和 3，第二学期上课程 4 ，第三学期上课程 1 ，第四学期上课程 5 。</span><br></pre></td></tr></table></figure>

<p><strong>示例 3：</strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：n &#x3D; 11, dependencies &#x3D; [], k &#x3D; 2</span><br><span class="line">输出：6</span><br></pre></td></tr></table></figure>

<p><strong>提示：</strong></p>
<ul>
<li><code>1 &lt;= n &lt;= 15</code></li>
<li><code>1 &lt;= k &lt;= n</code></li>
<li><code>0 &lt;= dependencies.length &lt;= n * (n-1) / 2</code></li>
<li><code>dependencies[i].length == 2</code></li>
<li><code>1 &lt;= xi, yi &lt;= n</code></li>
<li><code>xi != yi</code></li>
<li>所有先修关系都是不同的，也就是说 <code>dependencies[i] != dependencies[j]</code> 。</li>
<li>题目输入的图是个有向无环图。</li>
</ul>
<h3 id="解答"><a href="#解答" class="headerlink" title="解答"></a>解答</h3><p>深度遍历 + 剪枝也能通过</p>
<p>思路很简单 就是深度遍历，在入度为0的课程个数大于K时直接遍历所有组合<br>上面的思路时间复杂度太高了，于是对于于遍历过程中的相同情况进行剪枝</p>
<p>都不知道<code>help = sum(2 ** i for i in is0)</code>其实就是状态压缩</p>
<p>实际上动态规划会更好一些，这题会有详细解答的更新。</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">import</span> copy</span><br><span class="line"><span class="keyword">from</span> itertools <span class="keyword">import</span> combinations</span><br><span class="line"></span><br><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>:</span></span><br><span class="line">    <span class="function"><span class="keyword">def</span> <span class="title">minNumberOfSemesters</span>(<span class="params">self, n: int, dependencies, k: int</span>) -&gt; int:</span></span><br><span class="line">        <span class="comment"># 遍历 + 剪枝</span></span><br><span class="line">        <span class="keyword">global</span> res</span><br><span class="line">        depend = [<span class="number">0</span> <span class="keyword">for</span> _ <span class="keyword">in</span> range(n)]</span><br><span class="line">        root = [[] <span class="keyword">for</span> _ <span class="keyword">in</span> range(n)]</span><br><span class="line">        ways = &#123;&#125;</span><br><span class="line">        res = <span class="number">16</span></span><br><span class="line"></span><br><span class="line">        <span class="keyword">for</span> a, b <span class="keyword">in</span> dependencies:</span><br><span class="line">            root[a<span class="number">-1</span>].append(b<span class="number">-1</span>)</span><br><span class="line">            depend[b<span class="number">-1</span>] += <span class="number">1</span></span><br><span class="line"></span><br><span class="line">        <span class="function"><span class="keyword">def</span> <span class="title">backtrack</span>(<span class="params">choice,step,k,n</span>):</span></span><br><span class="line"></span><br><span class="line">            <span class="keyword">global</span> res</span><br><span class="line">            <span class="keyword">if</span> max(choice) == <span class="number">-1</span>:</span><br><span class="line">                res = min(res,step)</span><br><span class="line">                <span class="keyword">return</span> <span class="number">0</span></span><br><span class="line">            step += <span class="number">1</span></span><br><span class="line">            tempnum = <span class="number">0</span></span><br><span class="line">            is0 = []</span><br><span class="line">            <span class="keyword">for</span> i <span class="keyword">in</span> range(len(choice)):</span><br><span class="line">                <span class="keyword">if</span> choice[i] == <span class="number">0</span>:</span><br><span class="line">                    tempnum += <span class="number">1</span></span><br><span class="line">                    is0.append(i)</span><br><span class="line"></span><br><span class="line">            help = sum(<span class="number">2</span> ** i <span class="keyword">for</span> i <span class="keyword">in</span> is0)</span><br><span class="line">            <span class="keyword">if</span> help <span class="keyword">in</span> ways <span class="keyword">and</span> ways[help] &lt;= step:</span><br><span class="line">                <span class="comment">#pass</span></span><br><span class="line">                <span class="keyword">return</span> <span class="number">0</span></span><br><span class="line">            <span class="keyword">else</span>:</span><br><span class="line">                ways[help] = step</span><br><span class="line"></span><br><span class="line">            <span class="keyword">if</span> tempnum &lt;= k:</span><br><span class="line">                <span class="keyword">for</span> i <span class="keyword">in</span> is0:</span><br><span class="line"></span><br><span class="line">                    choice[i] -= <span class="number">1</span></span><br><span class="line">                    <span class="keyword">for</span> j <span class="keyword">in</span> root[i]:</span><br><span class="line">                        choice[j] -= <span class="number">1</span></span><br><span class="line">                backtrack(choice,step,k,n)</span><br><span class="line">            <span class="keyword">else</span>:</span><br><span class="line">                <span class="keyword">for</span> temp2 <span class="keyword">in</span> combinations(is0,k):</span><br><span class="line">                    temp3 = copy.deepcopy(choice)</span><br><span class="line">                    <span class="keyword">for</span> i <span class="keyword">in</span> temp2:</span><br><span class="line">                        temp3[i] -= <span class="number">1</span></span><br><span class="line">                        <span class="keyword">for</span> j <span class="keyword">in</span> root[i]:</span><br><span class="line">                            temp3[j] -= <span class="number">1</span></span><br><span class="line">                    backtrack(temp3,step,k,n)</span><br><span class="line"></span><br><span class="line">        backtrack(depend,<span class="number">0</span>,k,n)</span><br><span class="line">        </span><br><span class="line">        <span class="keyword">return</span> res</span><br><span class="line"></span><br></pre></td></tr></table></figure>


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